4

In classical mechanics, the orbital angular momentum of a particle is defined as $\textbf{L}=\textbf{r}\times\textbf{p}$. This is zero in the rest frame of the particle where $\textbf{p}=0$.

Quantum mechanically, $\textbf{p}$ is an operator. So putting $\hat{\textbf{p}}=0$ in $\hat{\textbf{L}}=\hat{\textbf{r}} \times\hat{\textbf{p}}$ and claiming that the orbital angular momentum of a quantum particle is zero in its rest frame does not make sense. One must look at the value of $\hat{\textbf{L}}^2$ on the "wavefunction in the rest frame" of the particle.

How does one find the wavefunction of a particle in its rest frame?

SRS
  • 27,790
  • 13
  • 115
  • 365

1 Answers1

6

The rest-frame wavefunction $\psi(\boldsymbol x,t)$ is the one such that $$ \boldsymbol 0\equiv\langle \boldsymbol p\rangle=\int_{\mathbb R^3}\psi^*(\boldsymbol x,t)(-i\boldsymbol \nabla)\psi(\boldsymbol x,t)\ \mathrm d\boldsymbol x $$

If $\boldsymbol k\equiv\langle \boldsymbol p\rangle$ is non-zero, we just need to redefine the wave-function: $$ \psi(\boldsymbol x,t)\to\mathrm e^{-i\boldsymbol k\cdot\boldsymbol x}\psi(\boldsymbol x,t) $$ which satisfies $\langle\boldsymbol p\rangle\equiv \boldsymbol 0$ by construction. This is just a translation in momentum space, $$ \tilde\psi(\boldsymbol p,t)\to \tilde\psi(\boldsymbol p-\boldsymbol k,t) $$ which obviously has zero mean.

More generally, if you have a system of many particles, the rest-frame of the system is, by definition, the one where $\langle\boldsymbol p\rangle\equiv\boldsymbol 0$, where $\boldsymbol p$ denotes the total linear momentum: $$ \boldsymbol p=\sum_i \boldsymbol p_i $$

AccidentalFourierTransform
  • 56,647